Chapter 9: Problem 49
In Problems 49-56, find the eigenvalues \(\lambda_{1}\) and \(\lambda_{2}\) and corresponding eigenvectors \(\mathbf{v}_{1}\) and \(\mathbf{v}_{2}\) for each matrix A. Determine the equations of the lines through the origin in the direction of the eigenvectors \(\mathbf{v}_{1}\) and \(\mathbf{v}_{2}\), and graph the lines together with the eigenvectors \(\mathbf{v}_{1}\) and \(\mathbf{v}_{2}\) and the vectors \(A \mathbf{v}_{1}\) and \(A \mathbf{v}_{2}\). $$ A=\left[\begin{array}{rr} 2 & 3 \\ 0 & -1 \end{array}\right] $$
Short Answer
Step by step solution
Find the Characteristic Polynomial
Solve the Characteristic Equation
Find Eigenvectors for \(\lambda_1 = 2\)
Find Eigenvectors for \(\lambda_2 = -1\)
Equation of Lines through the Origin
Graphing the Vectors and Lines
Final Verification
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Characteristic Polynomial
- Replace \( A \) with \( A - \lambda I \) leading to \( \begin{bmatrix} 2-\lambda & 3 \ 0 & -1-\lambda \end{bmatrix} \).
- Calculate the determinant: \((2-\lambda)(-1-\lambda) - (3 \times 0) = \lambda^2 - \lambda - 2 \).
Eigenvectors
- For eigenvalue \( \lambda_1 = 2 \): \( (A - 2I)\mathbf{v}_1 = 0 \) simplifies to \( \begin{bmatrix} 0 & 3 \ 0 & -3 \end{bmatrix}\begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix} \), resulting in the eigenvector \( \mathbf{v}_1 = \begin{bmatrix} 1 \ 0 \end{bmatrix} \).
- For eigenvalue \( \lambda_2 = -1 \): \( (A + I)\mathbf{v}_2 = 0 \) leads to \( \begin{bmatrix} 3 & 3 \ 0 & 0 \end{bmatrix}\begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix} \), giving the eigenvector \( \mathbf{v}_2 = \begin{bmatrix} 1 \ -1 \end{bmatrix} \).
Linear Transformation
- The transformation of \( \mathbf{v}_1 = \begin{bmatrix} 1 \ 0 \end{bmatrix} \) by matrix \( A \) results in \( A\mathbf{v}_1 = \begin{bmatrix} 2 \ 0 \end{bmatrix} \), which is a stretching by factor 2. This shows that the transformation scales vectors on the x-axis.
- Similarly, the transformation of \( \mathbf{v}_2 = \begin{bmatrix} 1 \ -1 \end{bmatrix} \) results in \( A\mathbf{v}_2 = \begin{bmatrix} -1 \ 0 \end{bmatrix} \), indicating a direction change and magnitude scaling by -1 on the line \( y = -x \).